Question 214129


{{{-2x^2+50=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{ax^2+bx+c}}} where {{{a=-2}}}, {{{b=0}}}, and {{{c=50}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (0 +- sqrt( (0)^2-4(-2)(50) ))/(2(-2))}}} Plug in  {{{a=-2}}}, {{{b=0}}}, and {{{c=50}}}



{{{x = (0 +- sqrt( 0-4(-2)(50) ))/(2(-2))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--400 ))/(2(-2))}}} Multiply {{{4(-2)(50)}}} to get {{{-400}}}



{{{x = (0 +- sqrt( 0+400 ))/(2(-2))}}} Rewrite {{{sqrt(0--400)}}} as {{{sqrt(0+400)}}}



{{{x = (0 +- sqrt( 400 ))/(2(-2))}}} Add {{{0}}} to {{{400}}} to get {{{400}}}



{{{x = (0 +- sqrt( 400 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{x = (0 +- 20)/(-4)}}} Take the square root of {{{400}}} to get {{{20}}}. 



{{{x = (0 + 20)/(-4)}}} or {{{x = (0 - 20)/(-4)}}} Break up the expression. 



{{{x = (20)/(-4)}}} or {{{x =  (-20)/(-4)}}} Combine like terms. 



{{{x = -5}}} or {{{x = 5}}} Simplify. 



So the answers are {{{x = -5}}} or {{{x = 5}}}